1. Contemporary Engineering Economics, 4 th edition, © 2007 Variations of Present Worth Analysis Lecture No.17 Chapter 5 Contemporary Engineering Economics Copyright © 2006

2. Contemporary Engineering Economics, 4 th edition, © 2007 Future Worth Criterion  Given: Cash flows and MARR (i)  Find: The net equivalent worth at a specified period other than “present”, commonly the end of project life  Decision Rule: Accept the project if the equivalent worth is positive. $75,000 $24,400 $27,340 $55,760 0 12 3 Project life

3. Contemporary Engineering Economics, 4 th edition, © 2007 Example 5.6 Net Future Worth at the End of the Project

4. Contemporary Engineering Economics, 4 th edition, © 2007 Alternate Way of Computing the NFW

5. Contemporary Engineering Economics, 4 th edition, © 2007 ABC 1PeriodCash Flow 20($75,000) 31$24,400 42$27,340 53$55,760 6PW(15%)$3553.46 7FW(15%)$5,404.38 =FV(15%,3,0,-B6) Excel Solution:

6. Contemporary Engineering Economics, 4 th edition, © 2007 Solving Example 5.6 with Cash Flow Analyzer Net Present Worth Net Future Worth Payback period Project Cash Flows

7. Contemporary Engineering Economics, 4 th edition, © 2007 Example 5.7 Future Equivalent at an Intermediate Time

8. Contemporary Engineering Economics, 4 th edition, © 2007 Built a hydroelectric plant using his personal savings of $800,000 Power generating capacity of 6 million kwhs Estimated annual power sales after taxes - $120,000 Expected service life of 50 years  Was Bracewell's $800,000 investment a wise one?  How long does he have to wait to recover his initial investment, and will he ever make a profit? Example 5.9 Project’s Service Life is Extremely Long

9. Contemporary Engineering Economics, 4 th edition, © 2007 Mr. Bracewell’s Hydroelectric Project

10. Contemporary Engineering Economics, 4 th edition, © 2007 How Would You Find P for a Perpetual Cash Flow Series, A?

11. Contemporary Engineering Economics, 4 th edition, © 2007 Capitalized Equivalent Worth  Principle: PW for a project with an annual receipt of A over infinite service life  Equation: CE(i) = A(P/A, i, ) = A/i A 0 P = CE(i)

12. Contemporary Engineering Economics, 4 th edition, © 2007 Practice Problem 10 $1,000 $2,000 P = CE (10%) = ? 0 Given: i = 10%, N = ∞ Find: P or CE (10%) ∞

13. Contemporary Engineering Economics, 4 th edition, © 2007 Solution 10 $1,000 $2,000 P = CE (10%) = ? 0 ∞

14. Contemporary Engineering Economics, 4 th edition, © 2007 A Bridge Construction Project  Construction cost = $2,000,000  Annual Maintenance cost = $50,000  Renovation cost = $500,000 every 15 years  Planning horizon = infinite period  Interest rate = 5%

15. Contemporary Engineering Economics, 4 th edition, © 2007 $500,000 $2,000,000 $50,000 0 15 30 4560 Years Cash Flow Diagram for the Bridge Construction Project

16. Contemporary Engineering Economics, 4 th edition, © 2007 Solution:  Construction Cost P 1 = $2,000,000  Maintenance Costs P 2 = $50,000/0.05 = $1,000,000  Renovation Costs P 3 = $500,000(P/F, 5%, 15) + $500,000(P/F, 5%, 30) + $500,000(P/F, 5%, 45) + $500,000(P/F, 5%, 60). = {$500,000(A/F, 5%, 15)}/0.05 = $463,423  Total Present Worth P = P 1 + P 2 + P 3 = $3,463,423

17. Contemporary Engineering Economics, 4 th edition, © 2007 Alternate way to calculate P 3  Concept: Find the effective interest rate per payment period  Effective interest rate for a 15-year cycle i = (1 + 0.05) 15 - 1 = 107.893%  Capitalized equivalent worth P 3 = $500,000/1.07893 = $463,423 15 304560 0 $500,000